Glycolysis, Link Reaction, Krebs Cycle & Electron Transport Chain ATP Yield

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Glycolysis (Cytoplasm: Anaerobic)

Occurs in the cytoplasm and requires no oxygen: 1. Phosphorylation: Glucose (6C) is phosphorylated using 2 molecules of ATP to form hexose bisphosphate (6C), which makes glucose more reactive and prevents it from diffusing out of the cell. 2. Lysis: Hexose bisphosphate splits into two molecules of Triose Phosphate (TP, 3C). 3. Oxidation: Each TP is oxidised to Pyruvate (3C) by removing hydrogen (accepted by 2 NAD⁺ → 2 reduced NAD) and synthesising 4 ATP by substrate-level phosphorylation. Net yield per glucose: 2 Pyruvate, 2 net ATP, and 2 reduced NAD.

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The Link Reaction (Mitochondrial Matrix)

Pyruvate (3C) is actively transported from the cytoplasm across the double mitochondrial membrane into the matrix. Pyruvate undergoes oxidative decarboxylation: one carbon is removed as CO₂, and hydrogen is removed to reduce NAD⁺ → reduced NAD, leaving a 2-carbon Acetyl group that combines with Coenzyme A to form Acetyl Coenzyme A (Acetyl CoA, 2C). Per glucose molecule (2× link reaction): 2 Acetyl CoA, 2 CO₂, 2 reduced NAD, and 0 ATP.

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The Krebs Cycle (Mitochondrial Matrix)

1. Acetyl CoA (2C) combines with a 4-carbon acceptor molecule (oxaloacetate) to form a 6-carbon molecule (citrate), releasing Coenzyme A for reuse. 2. In a series of enzyme-catalysed redox and decarboxylation steps, the 6C molecule loses 2 carbons as CO₂ to regenerate the 4C acceptor molecule. 3. Per single turn (per Acetyl CoA): produces 1 ATP (by substrate-level phosphorylation), 3 reduced NAD, 1 reduced FAD, and 2 CO₂. Per glucose molecule (2 turns): 2 ATP, 6 reduced NAD, 2 reduced FAD, and 4 CO₂.

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Oxidative Phosphorylation & Chemiosmosis (Cristae)

1. Reduced NAD and reduced FAD donate hydrogen atoms at the inner mitochondrial membrane (cristae), which split into protons (H⁺) and electrons (e⁻). 2. Electrons pass along the electron transport chain via redox reactions, releasing energy used by proton pumps to actively pump H⁺ from the matrix into the intermembrane space. 3. H⁺ accumulates in the intermembrane space, creating an electrochemical proton gradient (proton motive force). 4. H⁺ diffuses back into the matrix through ATP synthase channels (chemiosmosis), synthesising ATP from ADP + Pᵢ. 5. Oxygen acts as the terminal electron acceptor, combining with electrons and protons in the matrix to form water: ½O₂ + 2e⁻ + 2H⁺ → H₂O.

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Anaerobic Respiration (Lactate vs Ethanol Pathways)

In the absence of oxygen, the electron transport chain stops (no terminal electron acceptor), so the Link Reaction and Krebs Cycle halt due to a lack of oxidised NAD⁺ and FAD. To allow glycolysis (and its 2 net ATP yield) to continue, reduced NAD must be re-oxidised to NAD⁺ in the cytoplasm: In mammals/animals, Pyruvate (3C) accepts H from reduced NAD to form Lactate (3C) (reversible in the liver); in yeast/plants, Pyruvate (3C) is decarboxylated to ethanal (2C + CO₂), which accepts H from reduced NAD to form Ethanol (2C).

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Alternative Respiratory Substrates & Respiratory Quotient (RQ)

Lipids are hydrolysed into glycerol (phosphorylated to TP for glycolysis) and fatty acids (broken into many 2C fragments that enter the Krebs cycle as Acetyl CoA), yielding >2× more ATP per gram than carbohydrates because fatty acids contain many C-H bonds that produce vast quantities of reduced NAD/FAD. Proteins are hydrolysed to amino acids, deaminated (removing NH₂), and carbon skeletons enter as pyruvate or Krebs intermediates. Respiratory Quotient RQ = CO₂ produced / O₂ consumed (Carbohydrates = 1.0, Proteins ≈ 0.9, Lipids ≈ 0.7, Anaerobic > 1.0).

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1. Q1

   Describe the events of glycolysis and explain how a cell continues to produce ATP by glycolysis when no oxygen is available. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. In the cytoplasm, glucose (6C) is phosphorylated using two molecules of ATP to form hexose bisphosphate, making it more reactive [1 mark]. 2. Hexose bisphosphate splits into two molecules of 3-carbon triose phosphate (TP) [1 mark]. 3. Each TP is oxidised by removing hydrogen (which reduces 2 NAD to 2 reduced NAD) and converted into two molecules of pyruvate (3C), producing 4 ATP by substrate-level phosphorylation (a net gain of 2 ATP) [1 mark]. 4. In the absence of oxygen, pyruvate acts as a hydrogen acceptor and is reduced by reduced NAD (forming lactate in animals, or ethanol and CO₂ in yeast) [1 mark]. 5. This regenerates oxidised NAD, preventing the cell from running out of NAD so that the oxidation of triose phosphate in glycolysis can continue producing 2 net ATP [1 mark].

2. Q2

   Explain how reduced NAD produced during the Krebs cycle leads to the synthesis of ATP by oxidative phosphorylation in mitochondria. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Reduced NAD donates electrons and protons (hydrogen atoms) to the first electron carrier protein in the inner mitochondrial membrane (cristae), becoming oxidised back to NAD [1 mark]. 2. Electrons pass along a chain of electron carrier proteins at progressively lower energy levels in a series of redox reactions [1 mark]. 3. The energy released by the electrons is used to actively pump protons (H⁺ ions) from the mitochondrial matrix across the inner membrane into the intermembrane space, building up an electrochemical proton gradient [1 mark]. 4. Protons diffuse down their electrochemical gradient back into the matrix through proton channels in stalked ATP synthase enzymes (chemiosmosis), driving the phosphorylation of ADP + Pᵢ to form ATP [1 mark]. 5. Oxygen acts as the terminal electron acceptor in the matrix, combining with the electrons from the end of the chain and H⁺ ions to form water (½O₂ + 2e⁻ + 2H⁺ → H₂O) [1 mark].

3. Q3

   Dinitrophenol (DNP) is a chemical that makes the inner mitochondrial membrane leaky to protons (H⁺ ions). Explain why a person who ingests DNP experiences rapid weight loss (breakdown of fat stores) and a dangerous rise in body temperature. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Protons (H⁺) pumped into the intermembrane space diffuse back into the mitochondrial matrix directly across the leaky inner membrane instead of passing through ATP synthase channels, destroying the proton gradient [1 mark]. 2. Because chemiosmosis through ATP synthase is uncoupled from electron transport, very little ATP is synthesised by oxidative phosphorylation [1 mark]. 3. To meet cellular ATP demand, the body rapidly hydrolyses stored triglycerides and increases the rate of glycolysis, the Krebs cycle, and electron transport (causing weight loss) [1 mark]. 4. Because the potential energy of the proton gradient is no longer captured as chemical energy in ATP bonds, that energy is released directly as thermal energy (heat), causing fatal hyperthermia [1 mark].

4. Q4

   Explain why aerobic respiration of one molecule of glucose yields approximately 15 times more ATP than anaerobic respiration of one molecule of glucose. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Anaerobic respiration only completes glycolysis in the cytoplasm, yielding a net gain of just 2 ATP per glucose molecule via substrate-level phosphorylation [1 mark]. 2. In anaerobic respiration, most of the chemical potential energy remains locked inside the C-H and C-C bonds of lactate (or ethanol), which cannot be broken down further without oxygen [1 mark]. 3. In aerobic respiration, pyruvate enters the mitochondrial matrix and is completely oxidised to CO₂ and H₂O via the link reaction and Krebs cycle [1 mark]. 4. This produces 10 reduced NAD and 2 reduced FAD per glucose, which donate electrons to the electron transport chain to drive oxidative phosphorylation (~30-32 additional ATP) [1 mark].

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